## The challenge

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

**Note:** A leaf is a node with no children.

**Example 1:**

Input:root = [3,9,20,null,null,15,7]Output:2

**Example 2:**

Input:root = [2,null,3,null,4,null,5,null,6]Output:5

**Constraints:**

- The number of nodes in the tree is in the range
`[0, 10<sup>5</sup>]`

. `-1000 <= Node.val <= 1000`

## The definition for a binary tree node

```
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
```

## The solution in Java code

First we need to make sure that we catch our 2 edge cases:

- root is null
- root has no child nodes

Then we can check if left is null, it means that we can run the depth again, but with right, adding 1 level.

The same then applies to the right node.

Once this recursion is out, we make sure to use the built-in `Math.min()`

to do the same logic if not caught above:

- minimum value will be the left node under recursion
- compared to the right node under recursion

Making sure to increment our final output by 1, as we have not yet counted in the additional level yet.

```
class Solution {
public int minDepth(TreeNode root) {
if (root == null) return 0;
if (root.left == null && root.right == null) return 1;
if (root.left == null) {
return this.minDepth(root.right) + 1;
}
if (root.right == null) {
return this.minDepth(root.left) + 1;
}
return Math.min(this.minDepth(root.left), this.minDepth(root.right)) +1;
}
}
```